Question

The National Vaccine Information Center estimates that 88.5% of Americans have had chickenpox by the time they reach adulthood.
(a) Calculate the probability that exactly 98 out of 100 randomly sampled American adults had chickenpox during childhood.

(b) What is the probability that exactly 2 out of a new sample of 100 American adults have not had chickenpox in their childhood?

(c) What is the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox?

(d) What is the probability that at most 3 out of 10 randomly sampled American adults have not had chickenpox?

We now consider a random sample of 120 American adults.
(e) How many people in this sample would you expect to have had chickenpox in their childhood?

And with what standard deviation?

(f) What is the probability that 96 or fewer people in this sample have had chickenpox in their childhood? (round to 4 places)

Answer 1

Let , X be the number of Americans have a chickenpox by the time they reach adulthood.

Here ,

Therefore , the probability mass function of X is ,

; x=0,1,2,.....,n and q=1-p

= 0 ; otherwise

(a) Since , n=100 , p=0.885 , q=1-p=0.115

Now ,

Therefore ,

the probability that exactly 98 out of 100 randomly sampled American adults had chickenpox during childhood is 0.0004

(b) Since , n=100 , p=0.885 , q=1-p=0.115

Now , Required probability =

Therefore ,

the probability that exactly 2 out of a new sample of 100 American adults have not had chickenpox in their childhood is approximately 1.

(c) Since , n=10 , p= 0.885 , q=0.115

Now ,

Therefore, the probability that at least 1 out of 10 randomly sampled American adults have had chickenpox is approximately 1

(d) Since , n=10 , p= 0.885 , q=0.115

Now ,

Therefore, the probability that at most 3 out of 10 randomly sampled American adults have had chickenpox is approximately 0.0000

(e) Since , n=120 , p=0.885 , q=0.115

Mean=np=120*0.885=106

Standard deviation=

Therefore , 106 people in this sample would you expect to have had chickenpox in their childhood.
And with what standard deviation 3.4947

(f) Since , n=120 , p=0.885 , q=0.115

Now ,

Therefore ,  the probability that 96 or fewer people in this sample have had chickenpox in their childhood is 0.0048